Question

Using Runge-Kutta method, when $\frac{dy}{dx}=xy+y^{2}$, $y(0)=1$ is solved by taking $h=0.1$, then in the usual notation the value of $k_{1}$ is:

• A. 0.001
• B. 0.0001
• C. 0.01
• D. 0.1

Hint:

In RK method, always compute $f(x_0,y_0)$ first, then multiply by step size $h$.

Answer 1

The Correct Option is D

Concept: In Runge-Kutta method (order 4 or standard form), the first step is: \[ k_1 = h f(x_0,y_0) \] where \[ f(x,y)=xy+y^2 \]

Step 1: Write given values.
\[ x_0=0,\quad y_0=1,\quad h=0.1 \]

Step 2: Compute $f(x_0,y_0)$.
\[ f(0,1)=0\cdot 1 + 1^2 = 1 \]

Step 3: Compute $k_1$.
\[ k_1 = h f(x_0,y_0) \] \[ k_1 = 0.1 \times 1 = 0.1 \] \[ \boxed{k_1=0.1} \]